15-to-1 distillation protocol: why is the probability of error of the distilled state $35p^3$ and not $\binom{15}{3} p^3=455p^3$?

Question

The circuit performing the $15$-to-$1$ distillation of the magic state $T|+\rangle$ is shown below (figure taken from Litinski's paper). The first gates encode the $|+\rangle$ of the 15 qubit code. A logical $T$ is applied on this state (the gate is transversal for the $15$ qubit code), and we finally decode the state. We end up with an un-encoded magic state $T|+\rangle$.

The 15-qubit code is a distance 3 code: hence we can detect up to $2$ errors. Assuming the only noisy gates are the $T$ gates (yellow box in this image) and that they fail with a probability of error $p$, by post-selecting the states which return trivial syndromes, we can guarantee that the final state has a probability of error $O(p^3)$.

In his paper, Litinski says that the probability of error, at leading order, is $35p^3$. For me it should be $\binom{15}{3} p^3=455p^3$ (there are $\binom{15}{3}$ possibilities to have $3$ errors).

Where does the $35p^3$ comes from? Is it that some triplet of errors are actually harmless? If so, where is it explained?

Answer 1

It's because not every triplet is a failure. Most are detected.

A basic error is for one of the T gates to instead be Z*T. There are 15 T gates. Each of the 15 possible basic errors has a unique symptom. If you treat the four checks performed by the factory as the bits of a binary integer, this gives you an error 1 an error 2 and etc up to error 15. Because symptoms combine by xoring, the uniqueness of symptoms property implies every pair of errors has a non-zero combined symptom. This combined symptom can only be zero'd by adding a third error. That's why it's a distance 3 code.

Because each single error has a unique symptom, for a given pair there will always be exactly one unique third error to add that turns the pair into an undetected logical error. For example, starting from error 3 and error 14 you will need error $3 \oplus 14 = 13$ to finish a bad triplet. So the number of bad triplets is at most the number of pairs, of which there are ${15 \choose 2} = 105$. However, this overcounts by a factor of 3 because each failing triplet can be found redundantly via the ${3 \choose 2}$ different pairs within it. If the pair $a,b$ has the finisher $c$ forming $a,b,c$, then $a,c$ will have the finisher $b$ and $b,c$ will have the finisher $a$ so those three pairs all are telling you the same triplet.

So the total number of bad triplets is ${15 \choose 2} / {3 \choose 2} = 35$.